SOLUTION: I am having trouble grasping the setup of a distance word problem. Here's an example. She travels to work at 39 mph. Her return trip home she traveled 29 mph and it took 1/2 hour l
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Question 455119: I am having trouble grasping the setup of a distance word problem. Here's an example. She travels to work at 39 mph. Her return trip home she traveled 29 mph and it took 1/2 hour longer. What is the distance to work.
If someone could show me the example of how to set it up I think I can pick it up from there. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! She travels to work at 39 mph.
Her return trip home she traveled 29 mph and it took 1/2 hour longer.
What is the distance to work.
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Let d = the distance to work
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Write a time equation: time = dist/speed
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29 mph time - 39 mph time = .5 hrs - = .5
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Multiply it by (29*39) to clear the denominators, results
39d - 29d = .5(29*39)
10d = 565.5
d =
d = 56.55 mi to work
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See if this works out - =
1.95 - 1.45 = .5; confirms our solution
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Did this make sense to you?
